A 10-kg rock falls from a height of 8-m above the ground. What is the kinetic energy of the rock just before it hits the ground?

A small truck has gone over an embankment. The driver has a head injury and is confused, but is able to tell you that he is hau

vladimir2022 [97]

Answer:

"Emergency resource guidebook" would provide you with the best initial information about this chemical

Explanation:

This guidebook is for the first responders in case of the beginning of any dangerous goods or the hazardous materials transportation incident. It is used in firefighters, paramedics and for the police officers, in countries like Canada, Mexico and also in the United states when responding to these accident cases becomes delay. The book has different colours in it, for chemicals it will be green colour whereas the blue or the yellow colour shows the toxic inhalation materials. In TIH it has initial isolation distance and protective action distance information against the accident. The white is the host specific.

3 0

3 years ago

Suppose that a comet that was seen in 550 A.D. by Chinese astronomers was spotted again in year 1941. Assume the time between ob

pickupchik [31]

To solve the problem it is necessary to apply the concepts related to Kepler's third law as well as the calculation of distances in orbits with eccentricities.

Kepler's third law tells us that

$T^2 = \frac{4\pi^2}{GM}a^3$

Where

T= Period

G= Gravitational constant

M = Mass of the sun

a= The semimajor axis of the comet's orbit

The period in years would be given by

$T= 1941-550\\T= 1391y(\frac{31536000s}{1y})\\T=4.3866*10^{10}s$

PART A) Replacing the values to find a, we have

$a^3= \frac{T^2 GM}{4\pi^2}$

$a^3 = \frac{(4.3866*10^{10})^2(6.67*10^{-11})(1.989*10^{30})}{4\pi^2}$

$a^3 = 6.46632*10^{39}$

$a = 1.86303*10^{13}m$

Therefore the semimajor axis is $1.86303*10^{13}m$

PART B) If the semi-major axis a and the eccentricity e of an orbit are known, then the periapsis and apoapsis distances can be calculated by

$R = a(1-e)$

$R = 1.86303*10^{13}(1-0.997)$

$R= 5.58*10^{10}m$

7 0

3 years ago

Estimate the electric field at a point 2.40 cm perpendicular to the midpoint of a uniformly charged 2.00-m-long thin wire carryi

nadya68 [22]

Answer:

$E = 1.85*10^{12}\frac{N}{C}$

Explanation:

Hi!

The perpendicular distance 2.4cm, is much less than the distance to both endpoints of the wire, which is aprox 1m. Then the edge effect is negligible at this field point, and we can aproximate the wire as infinitely long.

The electric filed of an infinitely long wire is easy to calculate. Let's call z the axis along the wire. Because of its simmetry (translational and rotational), the electric field E must point in the radial direction, and it cannot depende on coordinate z. To calculate the field Gauss law is used, as seen in the image, with a cylindrical gaussian surface. The result is:

$E = \frac{\lambda}{2\pi \epsilon_0 r}\\\lambda=\text{charge per unit length}=\frac{4.95 \mu C}{2 m} = 2.475 \frac{C}{m}\\r=\text{perpendicular distance to wire}\\\epsilon_0=8.85*10^{-12}\frac{C^2}{Nm^2}$